Problem
GEO-B1-M02-P029 Reflections of the Orthocenter
In acute triangle \(ABC\), altitudes \(BD\) and \(CE\) meet at \(H\). The altitudes of triangle \(ADE\) meet at \(F\), and \(M\) is the midpoint of \(BC\). Prove that \(BH+CH\ge 2FM\).
C. Hint 1. Reflect \(H\) across lines \(AB\) and \(AC\).
D. Hint 2. Prove that \(F\) is the midpoint of the segment between the two reflected points.
E. Full solution.
Reflect \(H\) across \(AB\), and call the image \(C'\). Then \(E\) is the midpoint of \(HC'\), and \(BC'=BH\), because reflection preserves distances and point \(B\) lies on the axis of reflection.
Similarly, reflect \(H\) across \(AC\), and call the image \(B'\). Then \(D\) is the midpoint of \(HB'\), and \(CB'=CH\).
In triangle \(ADE\), line \(DF\) is perpendicular to \(AE\), and \(AE\subset AB\), so \(DF\parallel CE\). Similarly, \(EF\parallel BD\). Hence \(H E F D\) is a parallelogram.
In triangle \(HC'B'\), points \(E\) and \(D\) are the midpoints of sides \(HC'\) and \(HB'\). Therefore the parallelogram \(H E F D\) shows that \(F\) is the midpoint of side \(C'B'\).
Points \(M\) and \(F\) are the midpoints of \(BC\) and \(B'C'\). Thus \(2FM\le BC'+CB'\); this is the triangle inequality applied to the sum of the two vectors from \(B\) to \(C'\) and from \(C\) to \(B'\).
But \(BC'=BH\) and \(CB'=CH\). Therefore \(2FM\le BH+CH\), as required.
A. Source analysis. Main objects: an orthocenter, reflections, midpoints, and the triangle inequality. The obvious approach is to estimate \(FM\) directly, but the hidden observation is that after two reflections, \(F\) becomes the midpoint of a new segment. Number of key ideas: 4.
F. Difficulty justification. This is Level 7: a strong regional-style problem with reflections, parallelisms, and a final estimate.
G. Check. This is not a one-step exercise: it requires two reflections, recognition of a parallelogram, a midpoint of an auxiliary segment, and only then the inequality.